# Time Divided by Time is What?

Discussion in 'VHDL' started by rickman, Nov 18, 2012.

1. ### rickmanGuest

I thought a time quantity divided by a time quantity would be a real,
but it seems to be an integer. Can anyone confirm that? I can't find a
reference that discusses this.

Rick

rickman, Nov 18, 2012

2. ### KJGuest

Yes, time/time is an integer. If you need more precision than integer than you can do something like this as an example...

real(time/1ps)/1E12

Kevin Jennings

KJ, Nov 18, 2012

3. ### rickmanGuest

That's essentially what I did. Thanks for the confirmation.

My concern is that time has the potential of being much larger than an
integer can hold, but then it would have a lot more precision than a
real can hold either. I guess it is best to do time math in the time
domain until you need to convert it to something else, like a trig
function, e.g. sin(pi*1e-9*real((2*freq*now)/1ns)). But the trouble is
the intermediate integer can overflow if you aren't careful. Maybe
better is...

sin(2.0*pi*1e-9*real(freq)*real(now/1ns))

Rick

rickman, Nov 18, 2012
4. ### Paul UiterlindenGuest

To the previous two posters: between the number and the unit the space is
not optional.

vcom Message # 1207:
[13.2 Lexical elements, separators, and delimiters], line 130:
At least one separator is required between an identifier or an
abstract literal and an adjacent identifier or abstract literal.

Paul.

Paul Uiterlinden, Nov 19, 2012
5. ### valtih1978Guest

I think that physical value is internally an integer plus unit. So, you
do not loose anything when convert the time into integer. You might
afraid that integer is limited to 4 giga values. I think that the time
relies on `Universal integer` rather than STD.INTEGER and LRM is precise
at page 128: "Any physical type / The same type = Universal integer".
So, you do not loose anything again.

valtih1978, Nov 19, 2012
6. ### Paul UiterlindenGuest

Considering for example 5 ns / 3 ns, resulting in 1, I would say you lose
quite some precision.

Paul Uiterlinden, Nov 19, 2012
7. ### rickmanGuest

Thanks for the info. I'm often not online and so can't check the web.
I noticed that in the example code for my application they left out the
space, so that's how I typed it and it works ok. Doing a Google search
I found this discussion...

http://www.eda.org/isac/IRs-VHDL-87/IR0003.txt

"VHDL Issue Number: 0003
Classification: Examples, Notes, and Appendices
Language Version: VHDL-87
Summary: Examples in the LRM do not have a space
between the abstract literal and the
unit name of physical literals
"
....

"Description of Problem
----------------------

The fourth paragraph of section 13.2 of the LRM states that:

"... At least one separator is required between an identifier
or an abstract literal and an adjacent identifier or abstract
literal."

Thus a space is always required between the abstract literal and the
unit name in a physical literal. Nevertheless, many of the examples
in the LRM incorrectly include physical literals which contain no
space between the abstract literal and the unit name.
"

So it seems that at one time at least the LRM didn't practice what it
preached and so I expect all vendors will accept code that does not
include a space between the literal and the unit. Still, it's not a
good practice to follow. There is always the potential for a new tool
to be more strict and all your code would then break!

Rick

rickman, Nov 19, 2012